Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 50

Find the equation of the tangent line of the curve $\displaystyle y = x^4 + 2x^2 - x$ at Point $(1,2)$

Required:

Equation of the tangent line to the curve at $P(1,2)$

Solution:

Let $y' = m$ (slope)


$\begin{equation} \begin{aligned} \qquad y' = m =& \frac{d}{dx} (x^4) + \frac{d}{dx} (2x^2) - \frac{d}{dx} (x) && \text{} \\ \\ \qquad y' = m =& 4x^3 + 4x - 1 && \text{} \\ \\ \qquad m =& 4x^3 + 4x - 1 && \text{Substitute value of $x$ which is 1} \\ \\ \qquad m =& 4(1)^3 + 4 (1) - 1 && \text{} \\ \\ m =& 7 && \end{aligned} \end{equation}$


Solving for the equation of the tangent line:


$\begin{equation} \begin{aligned} \qquad y - y_1 =& m(x - x_1) && \text{Substitute the value of the slope $(m)$ and the given point} \\ \\ \qquad y - 2 =& 7 (x -1) && \text{Distribute 7 to the equation} \\ \\ \qquad y - 2 =& 7x - 7 && \text{Add 2 to each side} \\ \\ \qquad y =& 7x - 7 + 2 && \text{Combine like terms} \\ \\ \qquad y =& 7x - 5 && \text{Equation of the tangent line to the curve at $P(1, 2)$} \end{aligned} \end{equation}$